The magnitude of two vectors are 16 and 12 units respectively and the magnitude of their scalar product is `98sqrt2` units. The angle between the vectors would be

Options

A. 30°
B. 45°
C. 60°
D. 90°

Answer

To find the angle between two vectors given their magnitudes and the magnitude of their scalar product, we can follow these steps:

### Step 1: Identify the given values
- Magnitude of vector A, |A| = 16 units
- Magnitude of vector B, |B| = 12 units
- Magnitude of the scalar product, |A·B| = 98√2 units

### Step 2: Use the formula for the scalar product
The scalar product (dot product) of two vectors can be expressed as:
\[ A \cdot B = |A| |B| \cos \theta \]
Where θ is the angle between the two vectors.

### Step 3: Substitute the known values into the formula
We substitute the known values into the scalar product formula:
\[ 98\sqrt{2} = 16 \times 12 \times \cos \theta \]

### Step 4: Calculate the product of the magnitudes
Calculate \( 16 \times 12 \):
\[ 16 \times 12 = 192 \]

### Step 5: Rewrite the equation
Now we can rewrite the equation:
\[ 98\sqrt{2} = 192 \cos \theta \]

### Step 6: Solve for cos θ
To find cos θ, we rearrange the equation:
\[ \cos \theta = \frac{98\sqrt{2}}{192} \]

### Step 7: Simplify the fraction
Now we simplify the fraction:
\[ \cos \theta = \frac{98\sqrt{2}}{192} \approx 0.72 \]

### Step 8: Find the angle θ
To find θ, we take the inverse cosine:
\[ \theta = \cos^{-1}(0.72) \]

### Step 9: Calculate the angle
Using a calculator, we find:
\[ \theta \approx 44^\circ \]

### Step 10: Consider the negative value of cos θ
Since cosine can be positive or negative, we also consider:
\[ \theta = \cos^{-1}(-0.72) \]
This gives us:
\[ \theta \approx 136^\circ \]

### Conclusion
Thus, the angle between the two vectors can be either approximately \( 44^\circ \) or \( 136^\circ \). However, since the question might be looking for the acute angle, we take \( 44^\circ \) as the final answer.